What is the value of the following logarithm? $\log_{12} \left(\dfrac{1}{12}\right)$
Solution: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $12^{y} = \dfrac{1}{12}$ Any number raised to the power $-1$ is its reciprocal, so $12^{-1} = \dfrac{1}{12}$ and thus $\log_{12} \left(\dfrac{1}{12}\right) = -1$.